Steve314
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 Mar11 awarded Popular Question Dec18 comment If I raise a simplest-terms rational to an integer power, is the result automatically in simplest terms? @Bill - it turns out that Euclids Lemma was the bit I was missing for in that inductive proof, though it only really restates the shared-prime-factors proof anyway. Dec18 comment If I raise a simplest-terms rational to an integer power, is the result automatically in simplest terms? @Bill - OK, I hadn't thought about it that way. An obvious idea for an inductive proof (that might be more general) has been bugging me, but it seems on the one hand so trivial it's silly, on the other hand to be bugging me that somethings missing. Anyway, I should have time to work through some of these things in the next few days and it's so long since I made time for some math-related fun, so I'll get back to that. Dec18 comment If I raise a simplest-terms rational to an integer power, is the result automatically in simplest terms? @Bill - there's simpler things I took as obvious too, such as dealing with negative integer powers. One thing that's been niggling me, though, is why my brain insists on prime factors. Any shared factor between numerator and denominator will do. I guess it's because you can't express a number as a product of all its factors, as MPW showed above, so although it's not necessary, it's a nice form. Also, all factors of a number are products of some subbag (if that's a word) of its prime factors anyway, so in a sense all factors are still covered. Dec16 accepted If I raise a simplest-terms rational to an integer power, is the result automatically in simplest terms? Dec16 awarded Self-Learner Dec16 awarded Yearling Dec16 comment If I raise a simplest-terms rational to an integer power, is the result automatically in simplest terms? OK, that much makes sense, but I'm going to have to re-read this after some sleep. Dec16 comment If I raise a simplest-terms rational to an integer power, is the result automatically in simplest terms? Shouldn't $gcd(a,c)$ be smaller than both $a$ and $c$? I can see there should be a proof based on the greatest common divisor, if only because it's equal to the product of the shared prime factors, but something seems odd here. Should some of these GCDs be least common multiples? Dec16 answered If I raise a simplest-terms rational to an integer power, is the result automatically in simplest terms? Dec16 asked If I raise a simplest-terms rational to an integer power, is the result automatically in simplest terms? Dec8 awarded Caucus Jul21 answered Why is $\sin(d\Phi) = d\Phi$ where $d\Phi$ is very small? Jul2 awarded Curious Mar4 comment Are there more rational numbers than integers? @caters - possibly simpler way to put it - there's at least one way to biject rationals to/from positive integers. It doesn't matter that you can't determine which rational has count n without first checking all the rationals for counts 1..n-1 for possible duplicates. Having a non-recursive way to compute the mapping isn't a requirement. Mar4 comment Are there more rational numbers than integers? @caters - usual counting system for positive rationals - 0/1, 1/1, 2/1, 1/2, 3/1, 1/3, 4/1, 3/2, 2/3, 1/4, ... - all the rationals with numerator+denominator=1, then those with numerator+denominator=2, then those with numerator+denominator=3 and so on, but excluding those that are equivalent to rationals already counted. You can count as far as you want, though you can never finish counting - just like counting positive integers. To include negative rationals, just alternate positive then negative in the counting sequence. Mar4 comment Are there more rational numbers than integers? @caters - I think you're confusing "countable" with "finite". Countable means you can think of a scheme for counting that theoretically counts all of them - it doesn't mean the process of counting them has to be achievable. As for doing operations with rationals - actually, rationals are closed over all arithmetic operations, excluding division by zero of course. If start with a counted sequence of rationals 1..n, operations may result in rationals with a count larger than n, but that's just it - rationals are a countably infinite set. There's no requirement for any finite subset to be closed. Oct22 awarded Nice Question Sep27 accepted Is there a name for functions that are their own inverses Sep27 comment Is there a name for functions that are their own inverses Thanks - have to wait a few minutes before I can accept, but this is the answer.